What is the sum from n=1 to infinity of 4/(n!+4^n) ?

The sum from n=1 to infinity of 4/(n!+4^n) is converges

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sum from n=1 to infinity of 4/(n!+4^n)

n = 1 \sum \infty \frac{4}{n ! + 4 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{4}{n ! + 4 ^{n}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 1 \sum \infty \frac{1}{n ! + 4 ^{n}}

n \to \infty lim (n (\frac{\frac{1}{n ! + 4 ^{n}}}{\frac{1}{( n + 1 ) ! + 4 ^{n + 1}}} - 1)) = \infty

L > 1,

by the Raabe's test

= 4 converges

= converges

n = 1 \sum \infty \frac{9 n}{9 n + 1}

n = 2 \sum \infty (- 0.8)^{n}

n = 0 \sum \infty 7 (\frac{3}{4})^{n}

n = 0 \sum \infty \frac{5}{n}

n = 0 \sum \infty ∣ n ∣

What is the sum from n=1 to infinity of 4/(n!+4^n) ?
The sum from n=1 to infinity of 4/(n!+4^n) is converges